Ionization potential and electron affinity of fluorine are 17.42 and 3.45 eV atom-1 respectively. Electronegativity of fluorine will be approximately

Question

Ionization potential and electron affinity of fluorine are 17.42 and 3.45 eV atom-1 respectively. Electronegativity of fluorine will be approximately (A) 3 (B) 6 (C) 4 (D) 2

Tardigrade Admin · Accepted Answer

Electronegativity=0.374 ([IE+EA]/2)+0.17 =0.374 ([17.42+3.45]/2)+0.17=4.07 ≈4

Q. Ionization potential and electron affinity of fluorine are 17.42 and 3.45 eV $a t o m^{- 1}$ respectively. Electronegativity of fluorine will be approximately

3

6

4

2

Electronegativity $= 0.374 \frac{[ I E + E A ]}{2} + 0.17$
$= 0.374 \frac{[ 17.42 + 3.45 ]}{2} + 0.17 = 4.07 \approx 4$

Q. Ionization potential and electron affinity of fluorine are 17.42 and 3.45 eV atom−1 respectively. Electronegativity of fluorine will be approximately

3

6

4

2

Electronegativity=0.3742[IE+EA]​+0.17 =0.3742[17.42+3.45]​+0.17=4.07≈4

Q. Ionization potential and electron affinity of fluorine are 17.42 and 3.45 eV $a t o m^{- 1}$ respectively. Electronegativity of fluorine will be approximately

Electronegativity $= 0.374 \frac{[ I E + E A ]}{2} + 0.17$
$= 0.374 \frac{[ 17.42 + 3.45 ]}{2} + 0.17 = 4.07 \approx 4$